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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Giúp mình với!!!! CMR: a)2(a³+b³+c³-3abc)=(a+b+c)[(a-b)²+(b-c)²+(c-a)²] b)(a+b)(b+c)(c+a)+4abc=c(a+b)²+a(b+c)²+b(a+c)²

Tháng Một 31, 2023 Toán học 2 Comments 0 views

Toán Lớp 8: Giúp mình với!!!!
CMR:
a)2(a³+b³+c³-3abc)=(a+b+c)[(a-b)²+(b-c)²+(c-a)²]
b)(a+b)(b+c)(c+a)+4abc=c(a+b)²+a(b+c)²+b(a+c)²

Comments ( 2 )

  1. cattien
    cattien
    31 Tháng Một, 2023 at 6:34 sáng
    Reply
    $\begin{array}{l} 2\left( {{a^3} + {b^3} + {c^3} – 3abc} \right) = 2\left[ {{{\left( {a + b} \right)}^3} – 3ab\left( {a + b} \right) + {c^3} – 3abc} \right]\\  = 2\left[ {{{\left( {a + b + c} \right)}^3} – 3\left( {a + b} \right)c\left( {a + b + c} \right) – 3ab\left( {a + b + c} \right)} \right]\\  = 2\left( {a + b + c} \right)\left[ {{{\left( {a + b + c} \right)}^2} – 3\left( {a + b} \right)c – 3ab} \right]\\  = 2\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} – ab – bc – ca} \right)\\  = \left( {a + b + c} \right)\left( {2{a^2} + 2{b^2} + 2{c^2} – 2ab – 2bc – 2ca} \right)\\  = \left( {a + b + c} \right)\left[ {{{\left( {a – b} \right)}^2} + {{\left( {b – c} \right)}^2} + {{\left( {c – a} \right)}^2}} \right]\\ \end{array}$  
    b)
    $\begin{array}{l} c{\left( {a + b} \right)^2} + a{\left( {b + c} \right)^2} + b{\left( {a + c} \right)^2} – 4abc\\  = c\left( {{a^2} + 2ab + {b^2}} \right) + \left( {ab + ac} \right)\left( {b + c} \right) + b\left( {{a^2} + 2ac + {c^2}} \right) – 4abc\\  = \left( {ab + ac} \right)\left( {b + c} \right) + 2abc + {a^2}c + {b^2}c + {a^2}b + b{c^2} + 2abc – 4abc\\  = \left( {ab + ac} \right)\left( {b + c} \right) + bc\left( {b + c} \right) + {a^2}\left( {b + c} \right)\\  = \left( {b + c} \right)\left( {ab + ac + bc + {a^2}} \right)\\  = \left( {b + c} \right)\left( {c + a} \right)\left( {a + b} \right)\\  \Rightarrow \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) + 4abc = c{\left( {a + b} \right)^2} + a{\left( {b + c} \right)^2} + b{\left( {a + c} \right)^2} \end{array}$

  2. truongankimtrong
    truongankimtrong
    31 Tháng Một, 2023 at 6:34 sáng
    Reply
    a)VT=2(a³+b³+c³-3abc)
    =2(a³+b³+c³-3abc+3a²b+3ab²-3a²b-3ab²)
    =2[(a²+3a²b+3ab²+b³)+c³-(3a²b+3ab²+3abc)]
    =2[(a+b)³+c³-3ab(a+b+c)]
    =2{(a+b+c)[(a+b)²-c(a+b)+c²]-3ab(a+b+c)}
    =2(a+b+c)[(a+b)²-c(a+b)+c²-3ab]
    =2(a+b+c)(a²+2ab+b²-ac-bc+c²-3ab)
    =2(a+b+c)(a²+b²+c²-ab-bc-ac)
    =(a+b+c)(2a²+2b²+2c²-2ab-2bc-2ac)
    =(a+b+c)(a²+a²+b²+b²+c²+c²-2ab-2bc-2ac)
    =(a+b+c)[(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ac+c²)]
    =(a+b+c)[(a-b)²+(b-c)²+(a-c)²]
    =VP
    ⇒đpcm
    b)VP=c(a+b)²+a(b+c)²+b(a+c)²
    =c(a²+2ab+b²)+a(b²+2bc+c²)+b(a²+2ac+c²)
    =a²c+2abc+b²c+ab²+2abc+ac²+a²b+2abc+bc²
    =a²c+(2abc+2abc+2abc)+b²c+ab²+ac²+a²b+bc²
    =a²c+6abc+b²c+ab²+ac²+a²b+bc²
    =a²c+abc+abc+4abc+b²c+ab²+ac²+a²b+bc²
    =(abc+ac²+b²c+bc²)+(a²b+a²c+ab²+abc)+4abc
    =c(ab+ac+b²+bc)+a(ab+ac+b²+bc)+4abc
    =c[a(b+c)+b(b+c)]+a[a(b+c)+b(b+c)]+4abc
    =c(b+c)(a+b)+a(b+c)(a+b)+4abc
    =(a+b)(b+c)(c+a)+4abc
    =VT
    ⇒đpcm
     

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222-9+11+12:2*14+14 = ? ( )

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